Answer:
2< AC<8
Step-by-step explanation:
For a triangle to exits sum of two sides must always be greater than the the third side
⇒ AC< AB+ BC =5+3=8 in
⇒AC<8 in
Also, the difference of the two sides must be smaller than the length of third side
⇒ AC> | AB-BC| = |5-3| =2
AC> 2 in
therefore, we can say 2< AC<8. Hence, AC can takes 3,4,5,6 values
I guess so 35 ft or more than it
Answer:
Step-by-step explanation:
First confirm that x = 1 is one of the zeros.
f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26
f(1) = 2 - 14 + 38 - 26
f(1) = -12 + 38 = + 26
f(1) = 26 - 26
f(1) = 0
=========================
next perform a long division
x -1 || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26
2x^3 - 2x^2
===========
-12x^2 + 28x
-12x^2 +12x
==========
26x -26
26x - 26
========
0
Now you can factor 2x^2 - 12x + 26
2(x^2 - 6x + 13)
The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16
So you are going to get a complex result.
x = -(-6) +/- sqrt(-16)
=============
2
x = 3 +/- 2i
f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)
The zeros are
1
3 +/- 2i
Answer:
The eastbound train travels at 92 mph and the westbound train travels at 96 mph.
Step-by-step explanation:
We know that Velocity = distance / time
If the eastbound train travels at x miles per hour, the westbound train travels at x + 4 mph.
After 3.5 hours, they are 658 miles apart. If we solve for distance the formula given above we have: Distance = velocity x time. We have that the total distance between both trains is 658 miles.
Substituting the values of the problem and solving for x we have
Thus, the eastbound train travels at 92 mph and the westbound train travels at 96 mph.
The value of 3ab+5b-5 is 18
